A question regarding ❋166.44 in Whitehead & Russell's Principia Mathematica
It is very hard to "disentangle" this formula.
We have to start with the LHS of $✳166.44$ :
$\Sigma‘ \times P^{;} Q$.
The first step is to apply the "transformation" $✳166.1$ : $Q \times P = \Sigma‘P \underset{\overset{\textbf{.,}}{\ }}{\downarrow}^{;} Q$.
Note
In the following steps I'll use "$\downarrow$" in place of "down-arrow with dot & comma". Alas! "downarrow" is already used in PM for the couple $x \downarrow y$.
Applying $✳166.1$ to $\Sigma‘ \times P$ we get :
$\Sigma‘ P \downarrow^{;} \Sigma‘^{;} Q$.
Now we need $✳38.11$ to "play with" $\downarrow$ [see $✳38.11$ in vol.I, page 313 :
$x \downarrow‘ y = \downarrow y‘x = x \downarrow y$
and the example with "$\cap$" : $\alpha \cap \beta =\cap \beta‘ \alpha$. This is so because we can "read" the intersection of $\alpha$ and $\beta$ as a function $\cap \beta‘$ "applied to" $\alpha$.]
Assuming that we have correctly applied it (see the Addendum for details), we have :
$[\Sigma‘(\Sigma‘ P \downarrow)^{;}]^{;} Q$.
We finally apply $✳150.1$, which again is "quite simple" : $S^{;}Q = S \dagger Q$, to get :
$[\Sigma‘(\Sigma‘ P \downarrow) \dagger]^{;} Q$.
Addendum
We have to consider (thanks to George...) $✳116.44$, which is the cardinal counterpart of $✳166.44$ (please, note the same number after the dot ...) :
$(s‘k) \times \alpha = s‘(\times \alpha)‘‘k$
Forgetting the "switch" of LHS with RHS, it has clearly the same "form" of :
$\Sigma‘ \times P^{;} Q = (\Sigma‘ Q) \times P$.
The first step in the proof of $✳116.44$ is an application of $✳113.1$ :
$\beta \times \alpha = s‘ \alpha \underset{\overset{\textbf{,,}}{\ }}{\downarrow} ‘‘\beta$ Def ["downarrow with two commas"].
This, in turn, is the correlate of $✳166.1$ :
$Q \times P = \Sigma‘P \underset{\overset{\textbf{.,}}{\ }}{\downarrow}^{;} Q$ Def.
Now, back again to the proof of $✳116.44$ : we apply $✳113.1$ to the RHS [where $s‘$ is the "$\beta$" and $\alpha$ is the "$\alpha$"], to get :
$s‘(\times \alpha)‘‘k = s‘ \alpha \underset{\overset{\textbf{,,}}{\ }}{\downarrow}‘‘s‘ ‘‘k$.
Here we need some transformation of the above RHS [see the text of PM page 114 and George's answer] :
$s‘s‘‘ \alpha \underset{\overset{\textbf{,,}}{\ }}{\downarrow}‘ ‘‘k$
Now, we can exploit the analogy between $✳166.44$ and $✳116.44$, "mimicking" the steps of $✳116.44$ :
$s‘(\times \alpha)‘‘k = s‘ \alpha \underset{\overset{\textbf{,,}}{\ }}{\downarrow}‘‘s‘ ‘‘k = s‘s‘‘ \alpha \underset{\overset{\textbf{,,}}{\ }}{\downarrow}‘ ‘‘k$
with something like :
$\Sigma‘ \times P^{;} Q = \Sigma‘ P \underset{\overset{\textbf{.,}}{\ }}{\downarrow}^{;} \Sigma‘^{;} Q = \Sigma‘ \Sigma‘ P \underset{\overset{\textbf{.,}}{\ }}{\downarrow}^{;;}Q$
and then replace one of the two "$;$" with the "dagger" ($\dagger$), according to $✳150.1$.